Solve Simple Voltage Problems with V=I*R Formula"

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SUMMARY

The discussion focuses on solving voltage problems using the formula V=I*R, specifically in the context of diode behavior during positive and negative cycles. It establishes that during the positive cycle, the diode is reversed biased, acting as an open circuit, resulting in the output voltage being equal to the input voltage, calculated as 2.2 * √2. Conversely, during the negative cycle, the diode is forward biased, functioning as a short circuit with a forward voltage drop (Vd) of 0.7 V, leading to a maximum negative voltage of -0.7 V.

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  • Understanding of Ohm's Law and the formula V=I*R
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  • Familiarity with AC and DC voltage cycles
  • Basic circuit analysis skills
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http://img27.imageshack.us/img27/4937/66001732.jpg http://img4.imageshack.us/img4/2615/94339211.jpg

V=I*R
I think they are almost the same but I'm not sure as to how to solve them...
for the second question I think it was just -.7 because the negative value would only count the Vd, and make it negative.

But can someone tell me if this is right
 
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The output voltage is the same voltage that appears across the Diode. If we have some ideas about the voltgae across the diode we will know the output voltage.

During the positive cycle the diode is reversed biased and we can replace it with an open circuit.

http://img96.imageshack.us/img96/1327/opencircuit.jpg

In this case the output voltage is exactly the same as input voltage. So, the maximum positive output voltage is 2.2 * [tex]\sqrt{2}[/tex].

During the negative cycle the diode is forward biased and can be replaced by a short circuit (for ideal diode) or a voltage source (for somehow practical diode). In this problem we have Vd = 0.7 V so we replace the diode by a voltage source.

http://img198.imageshack.us/img198/8016/shortcircuitx.jpg

In this case the maximum negative voltgae is -0.7 V.
 
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